1 A student did not heat the empty crucible in Procedure Ste

1) A student did not heat the empty crucible in Procedure Step 2, and it retained some impurities or oils from previous use. So the mass of the “dirty” crucible was recorded, but all the impurities were subsequently burned off in the experiment. Will the reported mass of the product be too high, too low, or unaffected as a result of this error? Explain.

Empirical Formula Report Sheet Date: __________

1. Mass of crucible and lid (g) 28.6525

3. Mass of crucible, lid and magnesium (g) 28.8158

4. Mass of magnesium (g) 0.1633

5. Mass of crucible, lid and product (g)28.9204

7. Calculate the mass of product (g) 0.2679

8. Calculate the mass of oxygen (g) in the product 0.1046

9. Calculate the mass percent of magnesium and oxygen in the product ______% Mg ______% O 10. Calculate the moles of magnesium (show your calculation)

11. Calculate the moles of oxygen (O) in the product (show your calculation)

12. What is the empirical formula of the compound?

Solution

1) If a student did not heat the empty crucible in Procedure Step 2, and it retained some impurities or oils from previous use. So the mass of the “dirty” crucible was recorded (higher), but all the impurities were subsequently burned off in the experiment.

The reported mass of the product be : too low.

The mass of product is calculated from subtracting the empty crucible mass from the the final mass of crucible + product. Since the empty crucible was higher than the actual value, the subtracted mass would be lower.

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Empirical formula

9. mass% Mg in product = (0.1633/0.2679) x 100 = 60.96%

mass% oxygen in product = (0.1046/0.2679) x 100 = 39.04%

10. mass Mg = 0.1633 g

moles Mg = 0.1633 g/24.305 g/mol = 0.00672 mol

11. mass Oxygen = 0.1046 g

moles Oxygen = 01046 g/16 g/mol = 0.00654 mol

12. ratio moles (oxygen/magnesium) = 0.00654/0.00672 = 1

Empirical formula = MgO